a) Identify the kinematical constraint relating motions Y and X. The cable does NOT slip on the pulley. For items (c) & (e-f-g) use

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1 EAMPLE PROBLEM for MEEN 363 SPRING 6 Objectives: a) To erive EOMS of a DOF system b) To unerstan concept of static equilibrium c) To learn the correct usage of physical units (US system) ) To calculate natural frequencies an natural moe shapes e) To preict the response of a system using moal coorinates, case: constant amplitue loa f) To etermine the final position of the system once transient effects have isappeare g) To learn how to combine mathematical statements with explanatory sentences. Car must pull a heavy block stuck in a hollow an eep mining sha. The front en of car is tie to a big tree with a cable of stiffness K. A flexible cable of stiffness K is connecte to an inextensible cable that in turn, with a pulley system, is connecte to the block. The motorcar engine will rive the car upwar with force F (t) known. The amping coefficient (C) represents the viscous rag K between the block an sha walls. In the figure, Y enote the static equilibrium position F (SEP) of the system. M Y SEP means no motion of car an block, an engine turne off, F. Thus, at the SEP springs K K an K are alreay eflecte. For example, spring must support 5% of the block weight θ (W ) as easily seen from the cable & pulley constraint. Next, spring must also evelop a static force to hol 5% of W plus a fraction of the car weight, i.e. W sin( ). This knowlege is BASIC, oes not require of M elaborate thinking or eriving lengthy equations. a) Ientify the kinematical constraint relating motions Y an. The cable oes NOT slip on the pulley. C b) Draw free boy iagrams for the car an block, label all forces an show their constitutive relation in terms of the motion coorinates, if applicable. c) Determine the static eflection (δ s ) of each spring element ) Derive EOMs for the car an block motion in terms of coorinates &. For items (c) & (e-f-g) use K 5 lb : W in : 5 lb C : 5 lb in K : 5 lb W in : lb θ : π 8 e) Fin the system (unampe) natural frequencies an moe shapes, i.e. solve for the system eigenvalues an eigenvectors. f) For F (t) lbf fin the (unampe) response of the system using moal analysis g) Fin the final or terminal position of the system, i.e. as t, what are an? In reality, stiffness K represents the stiffness of the cable wrappe on the pulley. Rarely elements are rigi Copyright Luis San Anrés (6)

2 FREE BODY iagrams an kinematic constraints Definitions: F s force from elastic cable connecte to tree F s force from elastic cable connecting car to cable on pulley T Tension on cable F D viscous rag force F engine force δ s, δ s are static eflections for spring an, respectively Kinematic constraint inextensible cable T δ T δy, hence δ δy Ext force F Assume state of motion to raw FBD > Y, > FsK (δ ) M W T K Y T TK (-Y) + ½ W T T M W FD C /t Copyright Luis San Anrés (6) 3

3 MEEN 363 EAMPLE DOF- ANALYSIS: car pulling a block ORIGIN : K 5 lb : W in : 5lb C : 5lb in K : 5 lb W in : lb θ : π 8 (a) kinematic constraint - inextensible cable W M : g W M : g lb M lb M 3.8 masses nee be expresse in lb.^/ for consistency in EOM The cable length is constant, thus l c l c + Y an the kinematic constraint follows as Y (b) Derive EOMS: Assume a state of motion with -Y>, > motorcar pulls block Block of mass M: From the FBD iagram, with >, an apply Newton's n law to obtain: M W F Damper + T () where F Damper C () t t is the viscous rag force T K ( Y) + K δ s F s (3) T is the cable tension Force from spring. (-Y)>, an δs is the static eflection for spring. Car of mass M: From the FBD iagram, with >Y, apply Newton's n law to obtain: M W sin( θ) + F () t + F s T (4) where F(t) is engine force t F s K δ s (5) is the force from cable connecte to tree (c) Before proceeing - fin the springs' static eflection. For statics, assume NO motion, hence no engine force an since Y means static equilibrium position: an equations () an (4) reuce to W T : an T K δ s F s W δ s : K δ s.5 in F s T + W sin θ an F s : W sin θ + W δ s : F s K δ s.8 in (6) Note that the spring attache to the fixe point (a tree for example) must hol 5% of block weight an a fraction of the car weight. It shoul be easy for you to erive the results in Eq. (6) w/o the ai of the complete equations () an (4), i.e. by simple statics.

4 () Derive EOMs for the -DOF system: Back to DYNAMICS substitute T an Famper into () to obtain M W F Damper + t M W C + K ( ) + K δ s t t T () an since K δ s W M C t t + K ( ) BLOCK of MASS M: M + C + 4K K t t (7) substitute Fs an T into (5) M W t sin( θ) + F () t + F s T (5) M W t M W t sin( θ) sin( θ) + F () t K δ s + K ( Y) + K δ s + F () t K δ s + K ( ) + K δ s Sub Y M W t sin( θ) + F () t K δ s + K K ( ) W from statics: K δ s W W sin( θ) + F s M F () t K K ( ) t cancelling terms, i.e. applying statics BLOCK of MASS M M + ( K + K ) K t F () t (8) Eqns. (7) an (8) are the esire equations of motion for the cart an block. Please recall that motions are from the static equilibrium position In matrix form, the EOMs are: M M t + K + K K K 4K + C t F () t (9) Note that mass & stiffness matrices are symmetric.the amping matrix is NOT

5 () Fin natural frequencies an natural moe shapes of UNDAMPED system. Disregaing amping, an letting the force F, eq. (9) becomes M M t + K + K K K 4K () Let a cos ωt a cos ωt () Substitution of () into () gives K + K M ω K K 4K M ω a a cos( ωt) cancel cos(ωt) since <> for all times The homogeneous system of eqns, K + K M ω K K 4K M ω a a () has a non-trivial solution if the eterminant of the system of equations equals zero, i.e. if K K Δω + M ω 4K M ω 4K Let λ ω an expaning the proucts in the eqn. above M 4 λ M M λ K + K + 4K M + K + K K 4K Let: aλ + bλ + c with: a : M M M b : K + K + 4K M IT IS MOST IMPORTANT HERE TO USE THE RIGHT PHYSICAL UNITS FOR MASS. 4 c : K + K K 4K a lb 4 b 5. 8 lb c 5.76 lb NOTE That the coefficients a,b,c have consistent physical units. That is, since the physical unit for λ is (/^), the physical units of the eterminant are (lbf/)^

6 The roots (eigenvalues) of the characteristic equation are.5 b b 4a c b + b 4a c λ : λ : a a an the natural frequencies are: ω ( λ ).5.5 : ω : λ.5 λ ω ra Now, fin the eigenvectors The two equations in () are linearly epenent. Thus, one cannot solve for a an a. Set a : arbitrarily; an from the first equation K + K M ω for ω a : K a for φ : a K + K M ω ω a : K a φ is the first eigenvector (natural moe).839 DOF () an DOF () move in phase, with > φ : a φ is the n eigenvector (natural moe).39 DOF () an DOF () move OUT of phase, with > CHECK: computational soware allows quick calculation of eigenvalues an eigenvectors Define M M : K : M K + K K K 4K as the mass an stiffness matrices Let MinvK : M K λ C : ( eigenvals( MinvK) ) φ C : eigenvec MinvK, λ φ C : eigenvec MinvK, λ λ C φ C φ C.3 foun from highest to lowest

7 These eigenvectors are ientical to the ones foun analytically, since φ C.839 an φ C φ C.39 φ C That is, the ratio of the on element to first element is fixe. The actual values are not important (f) Response in MODAL coorinates: Use the transformation q Ax f. Make moal matrix using eigenvectors A : augment φ, φ A f. check orthogonality property of natural moes M M : A T MA K M : A T KA M lb.75 6 M K M lb non-iagonal elements are very small non zero b/c of rounoff with computer f.3 efine moal masses an stiffnesses: M m : M M M m : M M,, K m : K M K m : K M,, check K m M m ra K m M m ra OK f.3 efine initial conitions: isplacements an velocities in moal coorinates At time ts, the system is at its static equilibrium position, hence the initial conitions are null isplacements an null velocities. Of course, the same applies to moal space, i.e. null initial isplacements an velocities for generality, efine: o : V o : Calculate inverse of A matrix A inv : A A

8 an in moal coorinates q o q o_ot q o : A inv o q o_ot : A inv V o q o f.4 Define moal force Q: A T F F o : lb F : F o lb Physical force vector f.5 moal response Q Using knowlege to solve 3 3 an since the initial conitions are null lb M mi q i t Both natural moes will be excite + K mi q Q i i i, q () t : Q cos ω t K m ( ) an q () t : Q cos ω t K m ( ) ω ra T large : π 6 ω arbitrary scale for plot.5 moal isplacements () q () t q () t q q t time () Note that there is no amping or attenuation of motions.

9 The response in physical coorinates, z an x, equals (from transformation xaq) t (): q () t + q () t A t :.839q () t.39q () t. The mean value of an are: isplacements () t () t () car -block t time () Q Q mean : + K m K m Q Q mean : K m K m mean mean will be use below Complicate response with two natural frequencies being excite. However, roughly ~ (as kinematics requires) (f) FINAL - terminal conition: With the MATh tools you have, We can NOT solve the problem fully since the amping matrix is NOT proportional. However, amping oes exist an aventually the system will achieve a new steay state conition. Since the applie force is a constant, In the limit as t approaches very, very large values t K + K K K 4K en en F o An solving this system of equations using Cramer's rule 4 4K Δ : K + K K F o 4K en : Δ K F o en : Δ en en en en mean mean Note that the graph of unampe perioic motions (t) an (t) shows oscillatory motions abut these terminal or en values.